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Solutions to the σk-Loewner-Nirenberg problem on

annuli are locally Lipschitz and not differentiable

YanYan Li ∗† and Luc Nguyen ‡

Dedicated to Alice Chang and Paul Yang on their 70th birthday

Abstract

We show for k ≥ 2 that the locally Lipschitz viscosity solution to the σk-

Loewner-Nirenberg problem on a given annulus a < |x| < b is C1, 1

k

loc in each

of a < |x| ≤√ab and

√ab ≤ |x| < b and has a jump in radial derivative

across |x| =√ab. Furthermore, the solution is not C

1,γloc for any γ > 1

k . Optimal

regularity for solutions to the σk-Yamabe problem on annuli with finite constant

boundary values is also established.

1 Introduction

Let Ω be a smooth bounded domain in Rn, n ≥ 3. For a positive C2 function u

defined on an open subset of Rn, let Au denote its conformal Hessian, namely

Au = − 2

n− 2u−n+2

n−2 ∇2u+2n

(n− 2)2u− 2n

n−2∇u⊗∇u− 2

(n− 2)2u− 2n

n−2 |∇u|2 I, (1.1)

and let λ(−Au) denote the eigenvalues of −Au. Note that Au, considered as a (0, 2)

tensor, is the Schouten curvature tensor of the metric u4

n−2 g where g is the Euclideanmetric.

∗Department of Mathematics, Rutgers University, Hill Center, Busch Campus, 110 Frelinghuysen

Road, Piscataway, NJ 08854, USA. Email: yyli@math.rutgers.edu.†Partially supported by NSF grant DMS-1501004.‡Mathematical Institute and St Edmund Hall, University of Oxford, Andrew Wiles

Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford OX2 6GG, UK. Email:

luc.nguyen@maths.ox.ac.uk.

1

For 1 ≤ k ≤ n, let σk : Rn → R denote k-th elementary symmetric function

σk(λ) =∑

i1<...<ik

λi1 . . . λik ,

and let Γk denote the cone Γk = λ = (λ1, . . . , λn) : σ1(λ) > 0, . . . , σk(λ) > 0.In [7, Theorem 1.1], it was shown that the σk-Loewner-Nirenberg problem

σk(λ(−Au)) = 2−k( n

k

)

, λ(−Au) ∈ Γk, u > 0 in Ω, (1.2)

u(x) → ∞ as dist(x, ∂Ω) → 0. (1.3)

has a unique continuous viscosity solution u and such u belongs to C0,1loc (Ω). Further-

more, u satisfies

limd(x,∂Ω)→0

u(x)d(x, ∂Ω)−n−2

2 = C(n, k) ∈ (0,∞). (1.4)

Equation (1.2) is a fully nonlinear elliptic equation of the kind considered byCaffarelli, Nirenberg and Spruck [3]. We recall the following definition of viscositysolutions which follows Li [20, Definitions 1.1 and 1.1’] (see also [19]) where viscositysolutions were first considered in the study of nonlinear Yamabe problems.

Let

Sk :=

λ ∈ Γk

∣

∣σk(λ) ≥ 2−k( n

k

)

, (1.5)

Sk := Rn \

λ ∈ Γk

∣

∣

∣σk(λ) > 2−k

( nk

)

. (1.6)

Definition 1.1. Let Ω ⊂ Rn be an open set and 1 ≤ k ≤ n. We say that an upper

semi-continuous (a lower semi-continuous) function u : Ω → (0,∞) is a sub-solution(super-solution) to (1.2) in the viscosity sense, if for any x0 ∈ Ω, ϕ ∈ C2(Ω) satisfying(u− ϕ)(x0) = 0 and u− ϕ ≤ 0 (u− ϕ ≥ 0) near x0, there holds

λ(

− Aϕ(x0))

∈ Sk

(

λ(

−Aϕ(x0))

∈ Sk, respectively)

.

We say that a positive function u ∈ C0(Ω) satisfies (1.2) in the viscosity sense ifit is both a sub- and a super-solution to (1.2) in the viscosity sense.

Equation (1.2) satisfies the following comparison principle, which is a consequenceof the principle of propagation of touching points [23, Theorem 3.2]: If v and w are

2

viscosity sub-solution and super-solution of (1.2) and if v ≤ w near ∂Ω, then v ≤ win Ω; see [7, Proposition 2.2]. The above mentioned uniqueness result for (1.2)-(1.3)is a consequence of this comparison principle and the boundary estimate (1.4).

In the rest of this introduction, we assume that Ω is an annulus a < |x| < b ⊂ Rn

with 0 < a < b < ∞, unless otherwise stated. C2 radially symmetric solutions to (1.2)were classified by Chang, Han and Yang [5, Theorems 1 and 2]. As a consequence,when 2 ≤ k ≤ n, there is no C2 radially symmetric function satisfying (1.2)-(1.3).On the other hand, the aforementioned uniqueness result from [7, 23] implies thatthe solution u to (1.2)-(1.3) is radially symmetric (since u(R·) is also a solution forany orthogonal matrix R). Therefore, (1.2)-(1.3) has no C2 solutions.

Our first result improves on the above non-existence of C2 solutions to (1.2)-(1.3).

Theorem 1.2. Suppose that n ≥ 3 and Ω is a non-empty open subset of Rn. Thenthere exists no positive function u ∈ C2(Ω) such that λ(−Au) ∈ Γ2 in Ω and that

(Ω, u4

n−2 g) admits a smooth minimal immersion f : Σn−1 → Ω for some smoothcompact manifold Σn−1.

Theorem 1.2 bears some resemblance to a result of Schoen and Yau [28] on arelation between positive scalar curvature and stable minimal surfaces.

Noting that when u is radially symmetric, ∂Br0 is minimal with respect to u4

n−2 g

if and only if ddr

∣

∣

r=r0(r

n−2

2 u(r)) = 0, we obtain the following corollary with r0 being a

minimum point of rn−2

2 u(r).

Corollary 1.3. Suppose that n ≥ 3. Let Ω = a < |x| < b ⊂ Rn with 0 < a < b < ∞

be an annulus. Then there exists no radially symmetric positive function u ∈ C2(Ω)such that λ(−Au) ∈ Γ2 in Ω and u(x) → ∞ as x → ∂Ω.

Our next result shows that the locally Lipschitz solution u is not C1.

Theorem 1.4. Suppose that n ≥ 3 and 2 ≤ k ≤ n. Let Ω = a < |x| < b ⊂ Rn with

0 < a < b < ∞ be an annulus and u be the unique locally Lipschitz viscosity solutionto (1.2)-(1.3). Then u is radially symmetric, i.e. u(x) = u(|x|),(i) u is smooth in each of a < |x| <

√ab and

√ab < |x| < b,

(ii) u is C1, 1k but not C1,γ with γ > 1

kin each of a < |x| ≤

√ab and

√ab ≤ |x| <

b,

(iii) and the first radial derivative ∂ru jumps across |x| =√ab:

∂r ln u∣

∣

r=√ab

− = −n− 2√ab

and ∂r lnu∣

∣

r=√ab

+ = 0.

3

A related problem in manifold settings is to solve on a given closed Riemannianmanifold (M, g) the equation

σk

(

λ(

− Au

4n−2 g

))

= 2−k( n

k

)

, λ(

− Au

4n−2 g

)

∈ Γk, u > 0 in M, (1.7)

where Au

4n−2 g

is the so-called Schouten tensor of the metric u4

n−2 g,

Au

4n−2 g

= − 2

n− 2u−1∇2

gu+2n

(n− 2)2u−2du⊗ du− 2

(n− 2)2u− 2n

n−2 |∇gu|2g g + Ag,

and where λ(

−Au

4n−2 g

)

is the eigenvalue of −Au

4n−2 g

with respect to the metric u4

n−2 g.

Equations (1.2) and (1.7) are fully non-linear and non-uniformly elliptic equations ofHessian type, usually referred to as the σk-Yamabe equation in the ‘negative case’,which is a generalization of the Loewner-Nirenberg problem [24]. This equation andits variants have been studied in Chang, Han and Yang [5], Gonzalez, Li and Nguyen[7], Gurksy and Viaclovsky [12], Li and Sheng [18], Guan [9], Gursky, Streets andWarren [11], and Sui [30]. For further studies on the counterpart of (1.2) in thepositive case, see [4, 6, 10, 13, 16, 17, 20, 21, 29, 31, 32] and the references therein.

We observe the following result, which is essentially due to Gursky and Viaclovsky[12]. We provide in the appendix the detail for the piece which is not directly availablefrom [12].

Theorem 1.5. Suppose that n ≥ 3, 2 ≤ k ≤ n, and (Mn, g) is a compact Riemannianmanifold. If λ(−Ag) ∈ Γk on M , then (1.7) has a Lipschitz viscosity solution.

Here viscosity solution is defined analogously as in Definition 1.1.

Definition 1.6. Let (Mn, g) be a Riemannian manifold, 1 ≤ k ≤ n, and Sk and Sk

be given by (1.5) and (1.6). We say that an upper semi-continuous (a lower semi-continuous) function u : M → (0,∞) is a sub-solution (super-solution) to (1.7) inthe viscosity sense, if for any x0 ∈ M , ϕ ∈ C2(M) satisfying (u − ϕ)(x0) = 0 andu− ϕ ≤ 0 (u− ϕ ≥ 0) near x0, there holds

λ(

− Aϕ

4n−2 g

(x0))

∈ Sk

(

λ(

−Aϕ

4n−2 g

(x0))

∈ Sk, respectively)

.

We say that a positive function u ∈ C0(M) satisfies (1.7) in the viscosity sense ifit is both a sub- and a super-solution to (1.7) in the viscosity sense.

4

In both contexts, it is an interesting open problem to understand relevant condi-tions on Ω, or on (M, g), which would ensure that (1.2)-(1.3), or (1.7) respectively,admits a smooth solution. We make the following conjecture.

Conjecture 1.7. Suppose that n ≥ 3, 2 ≤ k ≤ n, and Ω ⊂ Rn is a bounded smooth

domain. Then the locally Lipschitz viscosity solution to (1.2)-(1.3) is smooth near∂Ω.

Some further questions are in order.

Question 1.8. Suppose that n ≥ 3, 2 ≤ k ≤ n, and Ω ⊂ Rn is a bounded smooth

domain. If (1.2)-(1.3) has a smooth sub-solution, must (1.2)-(1.3) have a smoothsolution?

Question 1.9. Suppose that n ≥ 3, 2 ≤ k ≤ n, and Ω ⊂ Rn is a smooth strictly

convex (non-empty) domain. Is the locally Lipschitz viscosity solution to (1.2)-(1.3)smooth?

If Ω is a ball, then the solution to (1.2)-(1.3) is smooth and corresponds to thePoincare metric.

Question 1.10. Suppose that n ≥ 3, 2 ≤ k ≤ n, and Ω = Ω2 \ Ω1 6= ∅ whereΩ1 ⋐ Ω2 ⊂ R

n are smooth bounded strictly convex domains. Is the locally Lipschitzviscosity solution to (1.2)-(1.3) C2?

In the case Ω1 and Ω2 are balls, Ω = Ω2\Ω1 is conformally equivalent to an annulus,and so, by Theorem 1.4, the solution to (1.2)-(1.3) is not C2. We believe that theanswer to the above question is negative. We indicate here how such statement maybe proved. In view of Theorem 1.2, it suffices to show that if u is a C2 solution to

(1.2)-(1.3), then (Ω, u4

n−2 g) admits a smooth (immersed) minimal hypersurface. Itis reasonable to expect, in view of known results in the case k = 1 (cf. [1, 25]) andestimate (1.4), that

d(x, ∂Ω)∣

∣

∣∇(

u(x)d(x, ∂Ω)n−2

2

)

∣

∣

∣→ 0 as d(x, ∂Ω) → 0.

If the above estimate holds for k ≥ 2, one has that, for small δ > 0, the hypersurfaces

Xδ = x ∈ Ω : d(x, ∂Ω) = δ are strictly mean-convex with respect to u4

n−2 g and thenormal pointing toward the region enclosed between these two hypersurfaces. Thesehypersurfaces can be used as barriers to construct a desired minimal hypersurface,at least for n ≤ 7. For example, in dimension n = 3, a result of Meeks and Yau[26, Theorem 7] (see also [14, Theorem 4.2]) implies that there exists a conformal

5

map f : S2 → Ωδ = x ∈ Ω : d(x, ∂Ω) > δ which minimizes area among all

homotopically nontrivial maps from S2 into Ωδ and either f is a conformal embedding

or a double covering map whose image is an embedded projective plane. Since allcompact surfaces in R

3 are orientable (see e.g. [27] or [15, Corollary 3.46]), f is aconformal embedding and so f(S2) is an embedded minimal sphere in (Ωδ, u

4g). Thiswill be followed up in a subsequent joint work with Jingang Xiong.

Question 1.11. Suppose that n ≥ 3, 2 ≤ k ≤ n, and (Mn, g) is a Riemannianmanifold such that λ(−Ag) ∈ Γk on M . Does (1.7) have a unique Lipschitz viscositysolution?

It is clear that (1.7) has at most one C2 solution by the maximum principle. In fact,if (1.7) has a C2 solution, then that solution is also the unique continuous viscositysolution in view of the strong maximum principle [2, Theorem 3.1]. Equivalently, if(1.7) has two viscosity solutions, then it has no C2 solution.

Question 1.12. Suppose that n ≥ 3 and 2 ≤ k ≤ n. Does there exist a Riemannianmanifold (Mn, g) such that λ(−Ag) ∈ Γk on M and (1.7) has a Lipschitz viscositysolution which is not C2?

Finally, we discuss the case where (1.3) is replaced by finite constant boundaryconditions

u||x|=a = c1 and u||x|=b = c2. (1.8)

We completely determine in the following theorem the regularity of the solutionto (1.2) and (1.8) depending on whether ln b

ais larger, equal to, or smaller than

2T (a, b, c1, c2) where

T (a, b, c1, c2) :=1

2

∫ 0

−|pb−pa|

1 + e−2η−2max(pa,pb)[

1− enη]1/k

−1/2

dη, (1.9)

pa = − 2n−2

ln c1 − ln a and pb = − 2n−2

ln c2 − ln b.

Theorem 1.13. Suppose that n ≥ 3 and 2 ≤ k ≤ n. Let Ω = a < |x| < b ⊂ Rn with

0 < a < b < ∞ be an annulus, and c1, c2 be two positive constants and let T (a, b, c1, c2)be given by (1.9). Then there exists a unique continuous viscosity solution to (1.2)and (1.8). Furthermore, u is radially symmetric, i.e. u(x) = u(|x|), and exactly oneof the following four alternatives holds.

Case 1: ln ba< 2T (a, b, c1, c2), and u is smooth in a ≤ |x| ≤ b,

6

Case 2: ln ba= 2T (a, b, c1, c2), b

n−2

2 c2 < an−2

2 c1, and u is smooth in a ≤ |x| < b, isC1, 1

k but not C1,γ with γ > 1kin a ≤ |x| ≤ b,

Case 3: ln ba= 2T (a, b, c1, c2), b

n−2

2 c2 > an−2

2 c1, and u is smooth in a < |x| ≤ b, isC1, 1

k but not C1,γ with γ > 1kin a ≤ |x| ≤ b,

Case 4: ln ba> 2T (a, b, c1, c2), and there is some m ∈ (a, b) such that

(i) u is smooth in each of a ≤ |x| < m and m < |x| ≤ b,(ii) u is C1, 1

k but not C1,γ with γ > 1kin each of a ≤ |x| ≤ m and

m ≤ |x| ≤ b,(iii) and the first radial derivative ∂ru jumps across |x| = m:

∂r ln u∣

∣

r=m− = −n− 2

mand ∂r ln u

∣

∣

r=m+ = 0.

Note that when ln ba= 2T (a, b, c1, c2), we have in view of the definition of T (a, b, c1, c2),

pa and pb that bn−2

2 c2 6= an−2

2 c1.

Remark 1.14. It is clear from Theorem 1.13 (in Cases 1–3) that if u is a C1 andradially symmetric solution to (1.2) in the viscosity sense in some open annulus Ωthen u ∈ C∞(Ω).

Remark 1.15. In Case 4, the exact value of m is

m =√ab exp

(1

2

∫ pa−p

pb−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη)

where p is the solution to

lnb

a=

∫ 0

pb−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη

+

∫ 0

pa−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη.

The following question is related to Question 1.8.

Question 1.16. Suppose that n ≥ 3, 2 ≤ k ≤ n and Ω = a < |x| < b ⊂ Rn with

0 < a < b < ∞. Does there exist constants c1, c2 with ln ba> 2T (a, b, c1, c2) such that

the problem (1.2) and (1.8) has a smooth sub-solution?

7

Recall that by Theorem 1.13, when ln ba> 2T (a, b, c1, c2), the problem (1.2) and

(1.8) has no smooth solution.For comparison, we recall here a result of Bo Guan [8] on the Dirichlet σk-Yamabe

problem in the so-called positive case which states that the existence of a smooth sub-solution implies the existence of a smooth solution.

We conclude the introduction with one more question.

Question 1.17. Let n ≥ 3, 2 ≤ k ≤ n and m 6= n − 1. Does there exist a smoothdomain Ω ⊂ R

n such that the locally Lipschitz solution to (1.2)-(1.3) is C2 away froma set Σ which has Hausdorff dimension m?

In Section 2, we prove all the results above except Theorem 1.5, whose proof isdone in the appendix. Theorem 1.2 is proved first in Subsection 2.1. We then provea lemma on the existence and uniqueness a non-standard boundary value problemfor the ODE related to (1.2) in Subsection 2.3 and use it to prove Theorem 1.4 inSubsection 2.4 and Theorem 1.13 in Subsection 2.5.

Acknowledgment

The authors would like to thank Matt Gursky and Zheng-Chao Han for stimulatingdiscussions. The authors are grateful to the referees for their very careful reading anduseful comments.

2 Proofs

2.1 Proof of Theorem 1.2

We will use the following lemma.

Lemma 2.1. For every symmetric n × n matrix M with λ(M) ∈ Γ2 and every unitvector m ∈ R

n, it holdsMij(δij −mimj) ≥ 0.

Proof. Using an orthogonal transformation, we may assume without loss of generalitythat M is diagonal with diagonal entries λ1 ≤ λ2 ≤ . . . ≤ λn. Then (λ1, . . . , λn) ∈ Γ2.It is well known that this implies λ1 + . . .+ λn−1 ≥ 0. Now as

Mij(δij −mimj) =

n∑

ℓ=1

λℓ −n

∑

ℓ=1

λℓm2ℓ ≥

n∑

ℓ=1

λℓ − λn

n∑

ℓ=1

m2ℓ =

n−1∑

ℓ=1

λℓ,

the conclusion follows.

8

We will use the following result on the mean curvatures of an immersed hyper-surface with respect to two conformal metrics. Let Ω ⊂ R

n be an open set. Equip

Ω with the Euclidean metric g and a conformal metric gu := u4

n−2 g where u is C2.Let f : Σn−1 → Ω be a smooth immersion of a compact manifold Σn−1 into Ω. Letu = u f , g = f ∗g. For every point p ∈ Σ, let HΣ(p) and HΣ,u(p) denote the meancurvature vectors associated to f at f(p) and with respect to g and gu, respectively.To dispel confusion, we note that, in our notation, the mean curvature is the trace ofthe second fundamental form. Note that if ν is a unit vector at f(p) normal to theimage of a small neighborhood of p, then

∂νu(f(p)) +n− 2

2(n− 1)g(HΣ(p), ν)u(f(p)) =

n− 2

2(n− 1)gu(HΣ,u(p), u

− 2

n−2 ν)un

n−2 . (2.1)

Lemma 2.2. Let Ω be an open subset of Rn, n ≥ 3, and f : Σn−1 → Ω be a smoothimmersion. If u ∈ C2(Ω) satisfies λ(−Au) ∈ Γ2 in Ω, then

∆gu+n− 2

4(n− 1)|HΣ,u|2guu

n+2

n−2 − n− 2

4(n− 1)|HΣ|2gu− 1

(n− 2)u|∇gu|2 ≥ 0 on Σ.

Proof. Fix some p ∈ Σn−1 and let ν be a unit vector at f(p) normal to the image ofa small neighborhood of p,. Recall that

Au = − 2

n− 2u−n+2

n−2

[

∇2u− n

(n− 2)u∇u⊗∇u+

1

n− 2|∇u|2 I

]

.

Applying Lemma 2.1 with M = −n−22u

n+2

n−2Au(f(p)) and m = ν yields

0 ≤ ∇i ∇ju(

δji − νiνj)

− 1

(n− 2)u|∇u|2 + n

(n− 2)u|∂νu|2.

This means

0 ≤ ∆gu+ g(HΣ, ν)∂νu f +n− 1

(n− 2)u|∂νu f |2 − 1

(n− 2)u|∇gu|2 on Σ.

Using (2.1) yields the conclusion.

Proof of Theorem 1.2. Suppose by contradiction that u ∈ C2(Ω) is such that λ(−Au) ∈Γ2 in Ω and (Ω, u

4

n−2 g) admits a smooth minimal immersion f : Σn−1 → Ω for somesmooth compact manifold Σn−1. Here we have renamed u in the statement of the

9

theorem as u for notational convenience. Let ν denote a continuous unit normal alongΣ. By Lemma 2.2, we have

∆gu− n− 2

4(n− 1)|HΣ|2gu− 1

(n− 2)u|∇gu|2 ≥ 0 on Σ.

Integrating over Σ, we thus have that HΣ ≡ 0 and u ≡ const on Σ. In particular,f : Σn−1 → Ω is a minimal immersion with respect to g. This is impossible as thereis no smooth minimal immersion in R

n with codimension one.

2.2 Preliminary ODE analysis

By the uniqueness result in [7, 23], the solutions u in Theorems 1.4 and 1.13 areradially symmetric, u(x) = u(r) where r = |x|.

As in [5, 32], we work on a round cylinder instead of Rn. Namely, let

t = ln r − 1

2ln(ab), ξ(t) = − 2

n− 2ln u(r)− ln r

so that u4

n−2 g = e−2ξ(dt2 + gSn−1). A direct computation gives that, at points whereu is twice differentiable,

σk(λ(−Au)) =(−1)k

2k−1

( n− 1k − 1

)

e2kξ(1− |ξ′|2)k−1[ξ′′ +n− 2k

2k(1− |ξ′|2)], (2.2)

where here and below ′ denotes differentiation with respect to t.Note that, for k ≥ 2, at points where u is twice differentiable, λ(−Au) ∈ Γk if

and only if σk(λ(−Au)) > 0 and |ξ′| > 1. Indeed, if σk(λ(−Au)) > 0 and |ξ′| > 1,then (2.2) implies σi(λ(−Au)) > 0 for 1 ≤ i ≤ k and so λ(−Au) ∈ Γk. Con-versely, if λ(−Au) ∈ Γk for some k ≥ 2, then σ1(λ(−Au)) > 0, σ2(λ(−Au)) > 0 andσk(λ(−Au)) > 0. Using (2.2), we see that the first two inequalities imply |ξ′| > 1.

By the same reasoning, we have, at points where u is twice differentiable, ifλ(−Au) ∈ Γ2, then |ξ′| ≥ 1.

We are thus led to study the differential equation

e2kξ(1− |ξ′|2)k−1[ξ′′ +n− 2k

2k(1− |ξ′|2)] = (−1)kn

2k. (2.3)

under the constraint that |ξ′| > 1.It is well known (see [5, 32]) that (2.3) has a first integral, namely

H(ξ, ξ′) := e(2k−n)ξ(1− |ξ′|2)k − (−1)ke−nξ is (locally) constant along C2 solutions.

10

Figure 1: The contours of H for k = 2, n = 7. Each radially symmetric viscositysolution to (1.2) lies on a single contour of H but avoid the shaded region, i.e. thedotted parts of the contours of H are excluded. Every smooth solution stays on oneside of the shaded region. Every non-smooth solution jumps (on one contour) fromthe part below the shaded region to the part above the shaded region at a singlenon-differentiable point.

A plot of the contours of H for k = 2, n = 7 is provided in Figure 1. See [5] for amore complete catalog.

Before moving on with the proofs of our results, we note the following statement.

Remark 2.3. As a consequence of Theorem 1.13, we have in fact that H(ξ, ξ′) is(locally) constant along viscosity solutions.

Proof. Fix a < b in the domain of u and apply Theorem 1.13 relative to the interval[a, b] with c1 = u(a) and c2 = u(b). If we are in cases 1–3, u is C2(a, b) and so H(ξ, ξ′)is constant in a < r < b. Suppose we are in case 4. We have that u is C2 in(a, m) ∪ (m, b) for some m and so H(ξ, ξ′) is constant in each of a < r < m andm < r < b. Also, as u is C1 in each of (a, m] and [m, b), we have by assertion (iii)in case 4 that

limr→m−

H(ξ(t), ξ′(t)) = H(ξ(t(m)), 1) = H(ξ(t(m)),−1) = limr→m+

H(ξ(t), ξ′(t)).

11

Hence H(ξ, ξ′) is also constant in a < r < b.

2.3 A lemma

Lemma 2.4. For any T > 0, there exists a unique classical solution ξ ∈ C∞(0, T ) ∩C

1, 1k

loc ([0, T )) to (2.3) in (0, T ) such that

limt→T−

ξ(t) = −∞, (2.4)

ξ′(0) = −1, ξ′(t) < −1 in (0, T ). (2.5)

Furthermore, for every γ ∈ ( 1k, 1], ξ /∈ C1,γ

loc ([0, T )).

Proof. We use ideas from [5].

Step 1: We start by collecting relevant facts from [5] about the classical solution ξp,qto (2.3) satisfying the initial condition ξp,q(0) = p and ξ′p,q(0) = q for p ∈ R, q ∈(−∞,−1) on its maximal interval of unique existence Ip,q = (T p,q, T p,q) ⊂ R.

Note that, since ξ′p,q(0) = q < −1, it follows from (2.3) that, for as long as ξp,qremains C2, ξ′p,q < −1. Thus, as H(ξp,q, ξ

′p,q) = H(p, q), we have in Ip,q that

ξ′p,q = −

1 + e−2ξp,q[

1 + (−1)kH(p, q)enξp,q]1/k

1/2

. (2.6)

By [5](Theorem 1, Cases II.2 and II.3 for even k and Theorem 2, Cases II.2 andII.3 for odd k), we have that T p,q is finite (corresponding to r+ being finite in thenotation of [5]). Furthermore,

limt→T

−p,q

ξp,q(t) = −∞. (2.7)

By (2.6) we thus have

T p,q =

∫ p

−∞

1 + e−2ξ[

1− |H(p, q)|enξ]1/k

−1/2

dξ. (2.8)

In this proof, we will only need to consider the case that (−1)kH(p, q) < 0. Thenby [5](Theorem 1, Case II.2 for even k and Theorem 2, Case II.2 for odd k), we havethat T p,q is also finite (corresponding to r− being finite in the notation of [5]) and

limt→T+

p,q

ξp,q(t) is finite, limt→T+

p,q

ξ′p,q(t) = −1, and limt→T+

p,q

ξ′′p,q(t) = −∞. (2.9)

12

Using (2.9) as well as the fact that H(ξp,q, ξ′p,q) = H(p, q) and ξp,q is decreasing,

we have in Ip,q that

ξp,q < limt→T+

p,q

ξp,q(t) = −1

nln |H(p, q)|. (2.10)

Differentiating (2.6), we see that, as t → T+p,q,

limt→T+

p,q

(t− T p,q)k−1

k ξ′′p,q(t) exists and belongs to (−∞, 0).

Thus ξp,q extends to a C1, 1

k function in a neighborhood of T p,q and ξp,q does not extendto a C1,γ function in any neighborhood of T p,q.

Before moving on to the next stage, we note that, in view of (2.6),

T p,q − T p,q =

∫ − 1

nln |H(p,q)|

−∞

1 + e−2ξ[

1− |H(p, q)|enξ]1/k

−1/2

dξ

=

∫ 0

−∞

1 + |H(p, q)| 2n e−2η[

1− enη]1/k

−1/2

dη. (2.11)

In particular, then length of Ip,q depends only on n, k and the value of H(p, q), ratherthan p and q themselves.

Step 2: We now define for each given p ∈ R a unique classical solution ξp to (2.3) insome maximal interval (0, Tp) satisfying ξp(0) = p, ξ′p(0) = −1 and ξ′p < −1 in (0, Tp).

It is clear that (−1)kH(p,−1) = −e−np < 0, and as ∂pH(p,−1) = (−1)kne−np 6= 0.By the implicit function theorem, there exist p and q < −1 such that H(p, q) =H(p,−1). Note that this implies

−e−np = (−1)kH(p, q) > −e−np and so p < p.

Letξp(t) = ξp,q(t+ T p,q) and Tp = T p,q − T p,q.

By Step 1, it is readily seen that ξp is smooth in (0, Tp), belongs to C1, 1

k

loc ([0, Tp)) and

13

no C1,γloc ([0, Tp)) with γ > 1

k, satisfies (2.3) and ξ′p < −1 in (0, Tp),

limt→T−

p

ξp(t) = −∞, (2.12)

ξp(0) = −1

nln |H(p, q)| = p, ξ′p(0) = −1, (2.13)

0 > ξ′′p,q(t) = O(t−k−1

k ) as t → 0+, (2.14)

and Tp =

∫ 0

−∞

1 + e−2η−2p[

1− enη]1/k

−1/2

dη. (2.15)

We claim that ξp is unique in the sense that if ξp ∈ C2(0, Tp) ∩ C1([0, Tp)) is a

solution to (2.3) in some maximal interval (0, Tp) satisfying ξp(0) = p, ξ′p(0) = −1

and ξ′p < −1 in (0, Tp), then Tp = Tp and ξp ≡ ξp. To see this, note that, ξp(t) =

ξξp(s),ξ′p(s)(t − s) for all t, s ∈ (0, Tp), since they both satisfy the same ODE in t and

agree up to first derivatives at t = s. By Step 1, ξp(t) → −∞ as t → T−p , and so, as

p < p and ξp(0) = p, there exists t0 ∈ (0, Tp) such that ξp(t0) = p. This implies that

H(p, ξ′p(t0)) = H(ξp, ξ′p) = H(p,−1) = H(p, q) and so ξ′p(t0) = q. We deduce that

t0 = −T p,q, Tp = Tp and ξp ≡ ξp,q(· − t0) ≡ ξp, as claimed.

Step 3: From (2.15), we see that, as a function of p, Tp is continuous and increasingand satisfies

limp→−∞

Tp = 0 and limp→∞

Tp = ∞.

Thus, for any given T > 0, there is a unique p(T ) such that Tp(T ) = T . The solutionξp(T ) to (2.3) gives the desired solution.

2.4 Proof of Theorem 1.4

Let T = 12ln b

aand t = ln r− 1

2ln(ab). We need to exhibit a function ξ : (−T, T ) → R

such that ξ is smooth in each of (0, T ) and (−T, 0), is C1, 1

k

loc but not C1,γloc for any γ > 1

k

in each of [0, T ) and (−T, 0], the function u defined by

u(r) = exp[

− n− 2

2

(

ξ(t) + ln r)]

solves (1.2)-(1.3) in a < r = |x| < b in the viscosity sense, and

(i) limt→±T ξ(t) = −∞,

14

(ii) ξ′(0−) = 1, ξ′(0+) = −1,

(iii) and |ξ′| > 1 in (−T, 0) ∪ (0, T ).

Indeed, let ξT : [0, T ) → R be the solution obtained in Lemma 2.4, and define

ξ(t) =

ξT (t) if 0 ≤ t < T,ξT (−t) if − T < t < 0.

It is clear that ξ satisfies all the listed requirements except for the statement that usatisfies (1.2) in the viscosity sense at r =

√ab. It remains to demonstrate, for any

given x0 with |x0| =√ab, that

(a) if ϕ is C2 near x0 and satisfies ϕ ≥ u near x0 and ϕ(x0) = u(x0), then λ(−Aϕ(x0)) ∈Γk and σk(λ(−Aϕ(x0))) ≥ 2−k

( nk

)

,

(b) and if ϕ is C2 near x0 and satisfies ϕ ≤ u near x0 and ϕ(x0) = u(x0), then either

λ(−Aϕ(x0)) /∈ Γk or λ(−Aϕ(x0)) ∈ Γk but σk(λ(−Aϕ(x0))) ≤ 2−k( nk

)

.

Without loss of generality, we may assume that x0 = (√ab, 0, . . . , 0).

Since ∂r ln u|r=√ab

− = −n−2√ab

< 0 = ∂r ln u|r=√ab

+ , there is no C2 function ϕ such

that ϕ ≥ u near x0 and ϕ(x0) = u(x0). Therefore (a) holds.Suppose now that ϕ is a C2 function such that ϕ ≤ u near x0 and ϕ(x0) = u(x0).

As u is radial, this implies that

−n− 2√ab

= ∂x1ln u|

r=√ab

− ≤ ∂x1lnϕ(x0) ≤ ∂x1

ln u|r=

√ab

+ = 0, (2.16)

∂x2lnϕ(x0) = . . . = ∂xn lnϕ(x0) = 0, (2.17)

(

∂xi∂xj

ϕ(x0)−1√ab

∂x1ϕ(x0)δij

)

2≤i,j≤n≤ 0. (2.18)

(For (2.18), note that the matrix on the left hand side is the Hessian of ϕ|∂B√ab

with respect to the metric induced on ∂B√ab by the Euclidean metric.) Now define

ϕ(x) = ϕ(|x|) = ϕ(|x|, 0, . . . , 0), t = ln r− 12ln(ab) and ξ(t) = − 2

n−2ln ϕ(r)− ln r. By

(2.16), we have that |dξdt(0)| ≤ 1 and so λ(−Aϕ(x0)) /∈ Γk.

Let O denote the diagonal matrix with diagonal entries 1,−1, . . . ,−1. Note that,in block form,

∇2ϕ(x0) +Ot∇2ϕ(x0)O = 2

(

∂2x1ϕ(x0) 00

(

∂xi∂xj

ϕ(x0))

2≤i,j≤n

)

.

15

Thus, by (2.18),

∇2ϕ(x0) +Ot∇2ϕ(x0)O ≤ 2

(

∂2x1ϕ(x0) 00 1√

ab∂x1

ϕ(x0)(δij)2≤i,j≤n

)

= 2∇2ϕ(x0).

Also, ϕ(x0) = ϕ(x0) and, in view of (2.17), ∇ϕ(x0) = ∇ϕ(x0). Hence

−Aϕ(x0)−OtAϕ(x0)O ≤ −2Aϕ(x0).

As λ(−Aϕ(x0)) /∈ Γk, it follows that λ(−Aϕ(x0)−OtAϕ(x0)O) /∈ Γk. Since the set ofmatrices with eigenvalues belonging to Γk is a convex cone (see e.g. [22, Lemma B.1]),we thus have that λ(−Aϕ(x0)) /∈ Γk or λ(−OtAϕ(x0)O) /∈ Γk. Since O is orthogonal,we deduce that λ(−Aϕ(x0)) /∈ Γk. We have verified (b) and thus completed the proof.

2.5 Proof of Theorem 1.13

As mentioned before, the uniqueness of solution follows from [7, 23]. We proceed toconstruct a radially symmetric solution with the indicated properties.

Let T = 12ln b

a, pa = − 2

n−2ln c1 − ln a and pb = − 2

n−2ln c2 − ln b. We will only

consider the case that pa ≥ pb (which is equivalent to bn−2

2 c2 ≥ an−2

2 c1). (The casepa < pb can be treated using an inversion about |x| =

√ab.) We then have

T (a, b, c1, c2) =1

2

∫ 0

pb−pa

1 + e−2η−2pa[

1− enη]1/k

−1/2

dη

=1

2

∫ pa

pb

1 + e−2ξ[

1− en(ξ−pa)]1/k

−1/2

dξ

(i) Suppose that T < T (a, b, c1, c2). We show that Case 1 holds.

Note thatH(pa,−1) = −(−1)ke−npa. Thus as T < T (a, b, c1, c2) and (−1)kH(pa, ·)is decreasing in (−∞,−1), we can find qa < −1 such that

T =1

2

∫ pa

pb

1 + e−2ξ[

1 + (−1)kH(pa, qa)enξ]1/k

−1/2

dξ. (2.19)

Recall the solution ξpa,qa to (2.3) considered in the proof of Lemma 2.4. By (2.8),we have that 2T < T pa,qa. We then deduce from (2.6) and (2.19) that

ξpa,qa(2T ) = pb.

16

It thus follows that ξ(t) = ξpa,qa(t + T ) is smooth in [−T, T ], satisfies (2.3) andξ′ < −1 in (−T, T ), as well as ξ(−T ) = pa and ξ(T ) = pb. Returning to u =exp

(

− n−22

(

ξ(ln r − 12ln(ab)) + ln r

)

we obtain the conclusion.

(ii) Suppose that T = T (a, b, c1, c2). We show that Case 3 holds.

Recalling the definition of T (a, b, c1, c2), we see that as T > 0, we have pa 6= pb.As pa ≥ pb, we have pa > pb. We can now follow the argument in (i) with ξpa,qareplaced by ξpa (defined in the proof of Lemma 2.4) to reach the conclusion. We omitthe details.

(iii) Suppose that T > T (a, b, c1, c2). We show that Case 4 holds.

In this case, we select p ≥ pa(≥ pb) such that

T =1

2

∫ 0

pb−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη

+1

2

∫ 0

pa−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη

Such p exists as the right hand side tends to T (a, b, c1, c2) when p → pa and divergesto ∞ as p → ∞. Recall the solution ξp defined in the proof of Lemma 2.4. Let

T+ =1

2

∫ 0

pb−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη

and

T− =1

2

∫ 0

pa−p

1 + e−2η−2p[

1− enη]1/k

−1/2

dη.

Then 2T± < Tp and the function ξp satisfies ξp(2T+) = pb and ξp(2T−) = pa.We then let

ξ(t) =

ξp(T+ − T− + t) if − T+ + T− ≤ t < T,ξp(−T+ + T− − t) if − T < t < −T+ + T−.

We can then proceed as in the proof of Theorem 1.4 to show that ξ is the desiredsolution.

17

A Appendix: Proof of Theorem 1.5

We abbreviate u4

n−2 g as gu. For small τ > 0, let

Aτgu = Agu + τtrgu(Agu)gu

= − 2

n− 2u−1 (∇2

gu+ τ∆gu g) +2n

(n− 2)2u−2du⊗ du− 2

(n− 2)2u− 2n

n−2 |∇gu|2g g

+ Ag +τ

2(n− 1)Rg g.

By [12, Theorem 1.4], we have for all sufficiently small τ > 0 that the problem

σk

(

λ(

−Aτguτ

))

= 2−k(

nk

)

, λ(

− Aτguτ

)

∈ Γk, uτ > 0 in M, (A.1)

has a unique smooth solution uτ . Furthermore, by [12, Propositions 3.2 and 4.1], thefamily uτ is bounded in C1(M) as τ → 0. (C2 bounds for uτ were also provedin [12], but these bounds are unbounded as τ → 0.) Hence, along some sequenceτi → 0, uτi converges uniformly to some u ∈ C0,1(M). To conclude, we show that uis a viscosity solution to (1.7).

For notational convenience, we rename uτi as ui.Fix some x ∈ M .

Step 1: We show that u is a sub-solution to (1.7) at x. More precisely, we show thatfor every ϕ ∈ C2(M) such that ϕ ≥ u on M and ϕ(x) = u(x) there holds that

λ(

−Agϕ(x))

∈

λ ∈ Γk

∣

∣

∣σk(λ) ≥ 2−k

(

nk

)

= Sk =: S. (A.2)

Here dg denotes the distance function of g and Bδ(x) denote the open geodesicball of radius δ and centered at x with respect to g. Fix some arbitrary small δ > 0so that ϕδ := ϕ+ δ dg(·, x)2 is C2 in Bδ(x).

Note that

ϕδ = ϕ+ δ3 ≥ u+ δ3 on ∂Bδ(x) and ϕδ(x) = u(x). (A.3)

Select xi,δ ∈ Bδ(x) such that

(ϕδ − ui)(xi,δ) = infBδ(x)

(ϕδ − ui) =: mi,δ.

18

By (A.3) and the uniform convergence of ui to u, we have that xi,δ ∈ Bδ(x). It followsthat

∇g(ϕδ − ui)(xi,δ) = 0, ∇2g(ϕδ − ui)(xi,δ) ≥ 0

and so−Aτi

gϕδ−mi,δ(xi,δ) ≥ −Aτi

gui(xi,δ).

Recalling (A.1), we hence have

λ(

− Aτigϕδ−mi,δ

(xi,δ))

∈ S. (A.4)

On the other hand, as x is the unique minimum point of ϕδ − u in Bδ(x), we havexi,δ → x and mi,δ → 0 as i → ∞. We can now pass i → ∞ in (A.4) to obtain

λ(

− Agϕδ(x)

)

∈ S.

Since δ is arbitrary, this proves (A.2) after sending δ → 0.

Step 2: We show that u is a super-solution to (1.7) at x, i.e. if ϕ ∈ C2(M) is suchthat ϕ ≤ u on M and ϕ(x) = u(x), then

λ(

− Agϕ(x))

∈ Rn \

λ ∈ Γk

∣

∣

∣σk(λ) > 2−k

( nk

)

= Sk =: S. (A.5)

The proof is analogous to that in Step 1. Fix some arbitrary small δ > 0 so thatϕδ := ϕ−δ = ϕ− δ dg(·, x)2 is C2 in Bδ(x). Clearly

ϕδ ≤ u− δ3 on ∂Bδ(x) and ϕδ(x) = u(x).

We next select xi,δ ∈ Bδ(x) such that

(ϕδ − ui)(xi,δ) = supBδ(x)

(ϕδ − ui) =: mi,δ.

As before, we have xi,δ ∈ Bδ(x), ∇g(ϕδ − ui)(xi,δ) = 0, ∇2g(ϕδ − ui)(xi,δ) ≤ 0 and

−Aτigϕδ−mi,δ

(xi,δ) ≤ −Aτigui

(xi,δ).

By (A.1), we hence have

λ(

− Aτigϕδ−mi,δ

(xi,δ))

∈ S. (A.6)

Also, as xi,δ → x and mi,δ → 0 as i → ∞, we can first pass i → ∞ and then δ → 0in (A.6) to reach (A.5).

19

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